MaxWong

May be math.not.maths's interpretation is correct, but I have another:

\(\dfrac{7x^3}{2}-\dfrac{4x^3}{2}\\ =\dfrac{3x^3}{2}\)

martindomexk: What is it? Can you explain briefly? If I think that it makes sense I am gonna join.

2(u + 5) = 12

2u + 10 = 12

2u = 2

u = 1

Yea. There is a song in Hong Kong which has the lyrics : "Side Angle Side, Side Side Side, Angle Side Angle, Angle Angle Side!!"

natural logarithm has a definition:

\(\text{If }\ln x = y\\ \text{Then }x = e^y\)

Posted a while before you posted this comment......

x = i is correct.

I have to sleep, I will give the answer for 2..

You probably know that: \(\ln z =\ln|z| +i \arg(z) \)

If you try \(\ln i\):

\(\ln i\\ = \ln |i| + i \arg(i)\\ = \ln 1 + i \arctan \dfrac{1}{0}\\ =\ln 1 + i \dfrac{\pi}{2}\\ = \dfrac{i\pi}{2}\)

If you try \(\ln(1+i)\):

\(\ln(1+i)\\ =\ln|1+i| + i \arg(1+i)\\ =\ln 2 + i \arctan{1}\\ =\ln 2 + \dfrac{i\pi}{4}\\ = \ln 2 + \dfrac{\ln i}{2}\)

Therefore x = i satisfies the equation.

Good proof. 

My own proof is mentioned above.

One of the correct answers, if I did not mention that x is not a real number.

Good proof.

My proof:

\(\text{Let the angle(arccos theta) be }\phi\\ \tan(\arccos\theta) \\ = \tan \phi\\ = \dfrac{\sqrt{1-\theta^2}}{\theta}\\ \text{Let arcsin theta be }\phi_1\\ \cot(\arcsin\theta)\\ =\dfrac{1}{\tan\phi_1}\\ =\tan(90^{\circ}-\phi_1)\\ =\tan \phi\\ =\tan(\arccos\theta)\)